Answer the question in pic.
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The counting of C(m,m) in the second problem :
--> Basic Counting Principle => C(n,m) to C(m,m) are altogether (n-m+1) brackets altogether that contains a certain C(m,m)
=> We get the proof by the basic property :
C( n+1 , m+1 ) = C( n , m ) + C( n , m+1 )
--> Basic Counting Principle => C(n,m) to C(m,m) are altogether (n-m+1) brackets altogether that contains a certain C(m,m)
=> We get the proof by the basic property :
C( n+1 , m+1 ) = C( n , m ) + C( n , m+1 )
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Yuichiro13:
Spartan !! Or should I say !! Agis Echestratus !
Can you delete all my answers to your Generics ? I wish to add them on your Google+ profile and skip here
No worries They will be deleted , u can answer there too!
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